(4z+5)(9-z)=0

Solution for (4z+5)(9-z)=0 equation:

(4z+5)(9-z)=0

We add all the numbers together, and all the variables

(4z+5)(-1z+9)=0

We multiply parentheses ..

(-4z^2+36z-5z+45)=0

We get rid of parentheses

-4z^2+36z-5z+45=0

We add all the numbers together, and all the variables

-4z^2+31z+45=0

a = -4; b = 31; c = +45;
Δ = b2-4ac
Δ = 312-4·(-4)·45
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-41}{2*-4}=\frac{-72}{-8}
=+9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+41}{2*-4}=\frac{10}{-8}
=-1+1/4
$